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MATHEMATICS CONTEST 2006 SOLUTIONS

PART I: 30 Minutes; NO CALCULATORS

Section A. Each correct answer is worth 1 point.

1. Evaluate: 20 + 06

Solution : a0 = 1 for any a ≠ 0, and 0b = 0 for any b ≠ 0. Therefore, 20 + 06 = 1 + 0 = 1.

 
2. The perimeter of a rectangle is 2006 inches. Its length is 800 inches. What is its width?

Solution: Perimeter equals 2L + 2W. If L = 800, 2W = 406, so W = 203 inches.

 
3. Express in simplest form :

Solution: The common denominator for these fractions is 120, so

 
4. Evaluate and write the answer in scientific notation: (280 × 106) ÷ (7 × 103)

Solution:

 
5. How many of these numbers are not divisible by 9? 27, 90, 108, 792, 162, 2006, 5445

Solution: The “brute force” approach would be to divide each of these numbers by 9, and see
which have a remainder. There is a faster approach of you know that a number is divisible by 9
if
(and only if) the sum of its digits is divisible by 9. For example, 108 is divisible by 9 because
1 + 0 + 8 = 9 is (obviously) divisible by 9. By this test, only 2006 is not divisible by 9, so the
answer is 1.

 
6. Evaluate: 12 + 3 × 4 − 6 + 8

Solution: Follow standard order of operations: 12 + 3 × 4 − 6 + 8 = 12 + 12 − 6 + 8 = 26.

 
7. The divisor is 11, the quotient is 3, and the remainder is 4. What is the dividend?

Solution: In the equation a ÷ b = c + d/b , often written as a = bc + d, the dividend is a, the
divisor b, the quotient c, and the remainder d. Therefore, the dividend is a = (11)(3)+4 = 37.

 
Section B. Each correct answer is worth 2 points.
 
8. Which one of the following expressions is not equal to the other four?
(A) 3x(x + 2)(x − 1) (B) −(1 − x)(3x2 + 6x) (C) 3(x3 + x2 − 2)
(D) (x + 2)(3x2 − 3x) (E) −6x + 3x2 + 3x3

Solution: C is different from the rest. This observation can be made in a number of ways; for
example, by noting that C does not have a common factor of x (that is, all the other expressions
equal 0 when x = 0).

 
9. Express as a fraction in simplest form:

Solution 1: If , then so 100x − x = 99x = 12.2. Therefore,

Solution 2: Note that

 
10. Factor completely: ax − by − bx + ay

Solution: Group like terms:
ax − by − bx + ay = ax + ay − by − bx = a(x + y) − b(x + y) = (a − b)(x + y).
Equivalently ,
ax − by − bx + ay = ax − bx − by + ay = x(a − b) + y(−b + a) = (a − b)(x + y).

 
11. The degree measure of an angle of a regular polygon cannot be
(A) 60 (B) 75 (C) 90 (D) 120 (E) 135

Solution: The answer is B: The total degree measure of an n-gon is 180(n − 2). (For example,
the angles in a triangle sum to 180°; in a quadrilateral, they sum to 360°, etc.) Therefore, in a
regular n-gon, an individual angle measure is This gives 60° when n = 6 (a regular
hexagon), 90° for n = 4 (a square), 120◦ for n = 3 (an equilateral triangle), and 135° for n = 8
(a regular octagon).

 
12. Write in simplest form as an exponential expression with a single base:

Solution: Use the rule of exponents that says

 
Section C. Each correct answer is worth 3 points.
 
13. If (x3 + 3x − 2) is divided by (x2 − 2), what is the remainder?

Solution 1: Carry out polynomial long division.

Solution 2: Any time you perform the division a ÷ b and get quotient q and remainder r , those
four quantities are related by the equation a = b · q + r . (See the solution to problem #7
on the previous page .) For polynomial division, the remainder r should be a polynomial of
degree smaller than the degree of b. So we want r (a first-degree polynomial) in the equation
x3 + 3x − 2 = (x2 − 2)q + r . In order to have an x3 term (but no x2 term) on the right side, q
must be x, so x3 + 3x − 2 = (x2 − 2)(x) + r, or r = (x3 + 3x − 2) − (x2 − 2x) = 5x − 2.

 
14. Express as a single radical in simplest form:

Solution:

 
15. (See figure). is tangent to at A. If BC = 3, CD = 3, OD = 2, and AB = 6, find
the length of the radius of .

Solution: Extend segment to intersect the circle at E. For any tangent segment (like ) and
secant segment (like), the lengths are related by the equation AB2 = (BC)(BE). Therefore,
BE = 12, and DE = 6.

Method 1: Extend into a diameter, intersecting
the circle at points F and G. A formula similar to the
one used above says that for any two secants (like
and ) which intersect inside the circle, we have
(CD)(DE) = (FD)(DG). Therefore, (3)(6) =
(r − 2)(r + 2), so r2 = 22, and r =

Method 2: Let H be the midpoint of , and note that Let x = OH. Then by the
Pythagorean theorem (applied to ∆OHD), x2 + 1.52 = 22, so x2 = 1.75. By the Pythagorean
theorem (applied to ∆OHC), x2 + 4.52 = r2, so r2 = 22, and r =

 
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